Determine $$\iint_S \textbf{A} \cdot d\textbf{S}$$ where $$\textbf{A}(x,y,z) = (xz^2,x^2y,xyz)$$ and S is the sphere $(x-1)^2+(y-2)^2+(z+3)^2=4$.
Attempted solution
Let's use Gauss' theorem. We obtain $\nabla \cdot \textbf{A} = x^2+z^2+xy$. A parametrization of the sphere is given by: $$\Phi(r,\theta,\phi)=\cases{x=r\sin(\phi)\cos(\theta)-1\\y=r\sin(\phi)\sin(\theta)-2\\z=r\cos(\phi)+3}\\$$ with $dV = r^2\sin(\phi)dr\,d\theta \,d\phi$. Furthermore we get: $$\textbf{A}(\Phi)=(r\sin(\phi)\cos(\theta)-1)^2+(r\cos(\phi)+3)^2+[r\sin(\phi)\cos(\theta)-1][(r\sin(\phi)\sin(\theta)-2]$$ Gauss theorem then gives: $$\begin{multline*}\iint_S \textbf{A} \cdot d\textbf{S} =\iiint_V \Big( r^2\sin(\phi)(r\sin(\phi)\cos(\theta)-1)^2+r^2\sin(\phi)(r\cos(\phi)+3)^2 \\+r^2\sin(\phi)[r\sin(\phi)\cos(\theta)-1][(r\sin(\phi)\sin(\theta)-2]\Big) \,dr\,d\theta\, d\phi\end{multline*}$$
Question
Is there a simpler way, because I'm not too keen on working with that beast?
Make a translational change of coordinates, namely, $$x = u + 1, \quad y = v + 2, \quad z = w - 3$$ so that the sphere $V$ is centered at the origin in the new coordinates. Then, $${\bf A}(u, v, w) = ((u + 1)(w - 3)^2, (u + 1)^2 (v + 2), (u + 1) (v + 2) (w - 3))$$ and so $$ \begin{align*} (\nabla \cdot {\bf A})(u, v, w) &= (w - 3)^2 + (u + 1)^2 + (u + 1)(v + 2) \\ &= u^2 + uv + w^2 + 4 u + v - 6 w + 12 . \end{align*} $$ If we integrate this expression over the sphere $V$, then by symmetry the $uv$, $u$, $v$, and $w$ terms do not contribute, and Gauss' Theorem gives $$\iint_S \nabla \cdot {\bf A} \,d{\bf S} = \iiint_V (u^2 + w^2 + 12) \,du\,dv\,dw.$$ Again by symmetry, the integrals of $u^2$ and $w^2$ over $V$ coincide, and our integral is equal to $$2 \iiint_V u^2 \,du\,dv\,dw + 12 \iiint_V \,du\,dv\,dw .$$ The first integral can be evaluated quickly using spherical coordinates, and, of course, the integral in the second term is just the volume of $V$.
Edit As user8268 points out in the comments, we can simplify the computation of the first integral in the last display equation by again appealing to symmetry: With the spherical coordinate change $r^2 = u^2 + v^2 + w^2$ in mind, we can rewrite the integral as $$\iiint_V u^2 \,du\,dv\,dw = \frac{1}{3} \iiint_V (u^2 + v^2 + w^2) \,du\,dv\,dw .$$